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Chapter 7: Problem 6

For exercises 1-66, simplify. $$ \frac{80 w^{3} z^{7}}{48 w^{9} z^{5}} $$

Short Answer

Expert verified
\[ \frac{5 z^{2}}{3 w^{6}} \]

Step by step solution

01

Factor out common terms in the numerator and the denominator

The given fraction is \ \ \ \ \ \ \ \ \[ \frac{80 w^{3} z^{7}}{48 w^{9} z^{5}} \] . \ \ \ \ Write the numerator and the denominator as products of primes and powers to identify common factors. This helps to cancel out the common terms.
02

Cancel out common factors

Divide both the numerator and the denominator by their greatest common divisor (GCD). Factor the coefficients: 80 and 48. \ \ \ \[ \frac{80}{48} = \frac{10}{6} = \frac{5}{3} \ (\because \text{GCD of 80 and 48 is 16}) \]. \ \ Now, simplify the variables: The exponents of \(w\) and \(z\) . \[ \frac{w^{3}}{w^{9}} = \frac{1}{w^{6}} \ (\because w^{3 - 9} = w^{-6}) \], and \[ \frac{z^{7}}{z^{5}} = z^{2} \ (\because z^{7 - 5} = z^{2}) \].
03

Write the simplified form

Now combine the simplified coefficients and variables: \[ \frac{5}{3} \times \frac{1}{w^{6}} \times z^{2} = \frac{5 z^{2}}{3 w^{6}} \]. \ So, the simplified expression is \[ \frac{5 z^{2}}{3 w^{6}} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Greatest Common Divisor
When simplifying algebraic fractions, finding the greatest common divisor (GCD) is a crucial step. The GCD is the largest number that divides both the numerator and the denominator without leaving a remainder. For example, in the fraction \(\frac{80}{48}\), the GCD of 80 and 48 is 16. This is because 16 is the highest number that can divide both 80 and 48. To find the GCD:
  • List the prime factors of each number
  • Identify the common factors
  • Multiply these common factors
In our case, the prime factors of 80 are 2, 2, 2, 2, and 5, while the prime factors of 48 are 2, 2, 2, and 3. The common factor here is 2, and multiplying it by itself three times gives us 16. Therefore, the GCD of 80 and 48 is 16.
Exponent Rules
Exponent rules are applied when simplifying expressions with variables that have exponents. These rules help in reducing complex expressions to simpler forms. Consider the exponents in the given problem. For the terms with \( w \) and \( z \), we divide their exponents:
  • For \( w \), the exponents are 3 and 9. Using the rule \( w^{a} / w^{b} = w^{a-b} \), we obtain \( w^{3-9} = w^{-6} \), which simplifies to \( \frac{1}{w^{6}} \).
  • For \( z \), the exponents are 7 and 5. Using the same rule, we get \( z^{7-5} = z^{2} \).
Knowing these exponent rules is essential for simplifying algebraic fractions.
Prime Factorization
Prime factorization involves breaking down a number into its prime factors. Prime numbers are numbers that have only two distinct positive divisors: 1 and itself. In the given exercise, breaking down the coefficients (80 and 48) into their prime factors can make it easier to simplify the fraction. For example:
  • Prime factors of 80 are 2, 2, 2, 2, and 5.
  • Prime factors of 48 are 2, 2, 2, 2, and 3.
By comparing these factors, we can easily identify the common ones and then divide both the numerator and denominator by these common prime factors. This step is necessary for simplifying the coefficients in a fraction, thereby reducing the fraction to its simplest form.

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Most popular questions from this chapter

$$ \text { Solve: } 0.75=\frac{k}{60} $$

For exercises 49-52, the formula \(C=\frac{P_{m} P_{i}}{T F}\) describes the cost of insurance, \(C\). Is the relationship of the given variables a direct variation or an inverse variation? $$ P_{i}, T \text {, and } F \text { are constant; the relationship of } C \text { and } P_{m} \text {. } $$

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When a car travels a fixed distance, the relationship between the speed of the car, \(x\), and the time it travels, \(y\), is an inverse variation. When the speed is \(\frac{48 \mathrm{mi}}{1 \mathrm{hr}}\), the time is \(0.75 \mathrm{hr}\). a. Find the constant of proportionality. Include the units of measurement. b. Write an equation that represents this relationship. c. Find the time in hours to travel this distance at a speed of \(\frac{80 \mathrm{mi}}{1 \mathrm{hr}}\). d. Change the time in part \(\mathrm{c}\) to minutes.
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